package ProjectEuler; /** * If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. * The sum of these multiples is 23. * *

Find the sum of all the multiples of 3 or 5 below 1000. * *

Link: https://projecteuler.net/problem=1 */ public class Problem01 { public static void main(String[] args) { int[][] testNumber = { {3, 0}, {4, 3}, {10, 23}, {1000, 233168}, {-1, 0} }; for (int[] ints : testNumber) { assert solution1(ints[0]) == ints[1]; assert solution2(ints[0]) == ints[1]; } } private static int solution1(int n) { int sum = 0; for (int i = 3; i < n; ++i) { if (i % 3 == 0 || i % 5 == 0) { sum += i; } } return sum; } private static int solution2(int n) { int sum = 0; int terms = (n - 1) / 3; sum += terms * (6 + (terms - 1) * 3) / 2; terms = (n - 1) / 5; sum += terms * (10 + (terms - 1) * 5) / 2; terms = (n - 1) / 15; sum -= terms * (30 + (terms - 1) * 15) / 2; return sum; } }